§5. Differentiable functions of several variables
283
J: (x,y) f---+ (h(x,y),h(x,y))
of class C 1 from U into V, whence a new composite function
p(x, y) = g [h (x, y), h(x, y)]
defined on U or, in condensed notation, p(z) = g[J(z)]. It is easy to see that
p is C 1 , and immediate to calculate its derivatives. These are obtained by
letting x vary while y remains constant, and vice versa; one finds oneself again
in the simplest situation just explained. On applying (1) either to x f--+ J(x, y),
or to y f--+ J(x, y), one then obtains
(21.13)
D1g[J(z)]Dd1(Z) + D 2 g[J(z)]D1h(z),
D1g[J(z)]D2h(z) + D 2 g[J(z)]D2h(z);
these formulae show that the derivatives of p are continuous, qed.
One can also start again from scratch using condensed notation as in
(12'). For z E U given and h E ]R2 small, Theorem 21 shows that
J(z + h)
p(z + h)
J(z) + DJ(z)h + o(h) = J(z) + k,
g[J(z) + k] = p(z) + Dg[J(z)]k + o(k).
Since k = DJ(z)h + o(h) = O(h) + o(h) = O(h), everything that is o(k) is
also o(h), and, on the other hand,
Dg[J(z)]k = Dg[J(z)]DJ(z)h + Dg[J(z)]o(h) = Dg[J(z)]DJ(z)h + o(h).
Consequently p is differentiable, with
Dp(z)h = Dg[J(z)]DJ(z)h,
the value of the linear map Dg[J(z)] at the vector DJ(z)h, itself the value of
the linear map DJ(z) at the vector h. In other words,
(21.13')
Dp(z) = Dg[J(z)] 0 DJ(z),
the composition ofthe linear maps Dg[J(z)] and D J(z); this is absolutely the
same proof - and the same formula - as for rule (D 4) of n° 15. One often
omits the sign 0 when dealing with linear maps, so writes (13) in the form
Dp(z) = Dg[J(z)]DJ(z).
We can of course make all this explicit in terms of h, ... , g2. We need only
know how to calculate the matrix of a product of linear maps as a function
of those of the factors. Whence, here,